- Displacement is the change in position of an object over time and is a vector quantity. It indicates both the distance and direction moved.
- Speed is the distance traveled per unit time and is a scalar quantity. It does not indicate direction.
- Velocity is speed with direction and is therefore a vector quantity. It indicates both how fast an object is moving as well as the direction of motion.
- Acceleration is the rate of change of velocity with time. It measures how velocity is changing and can therefore be positive, negative, or zero. Acceleration is a vector quantity.
2. Scalars Quantities
Scalars can be completely described by magnitude
(size)
Scalars can be added algebraically
They are expressed as positive or negative numbers and a
unit
Examples include :- mass, electric charge, distance,
speed, energy
3. Vector Quantities
Vectors need both a magnitude and a direction to
describe them (also a point of application)
When expressing vectors as a symbol, you need to
adopt a recognized notation
e.g. draw an arrow across top of the letter
They need to be added, subtracted and multiplied in
a special way
Examples :- velocity, weight, acceleration,
displacement, momentum, force
4. Addition and Subtraction
The Resultant (Net) is the vector that comes from
adding or subtracting a number of vectors
If vectors have the same or opposite directions the
addition can be done simply
same direction : add
opposite direction : subtract
5. Co-planar vectors
The addition of co-planar vectors that do not have
the same or opposite direction can be solved by using
scale drawings to get an accurate resultant
Or if an estimation is required, they can be drawn
roughly
or by Pythagoras’ theorem and trigonometry
Vectors can be represented by a straight line
segment with an arrow at the end.
6. Triangle of Vectors
Two vectors are added by drawing to scale and with the
correct direction the two vectors with the tail of one at
the tip of the other.
The resultant vector is the third side of the triangle and
the arrow head points in the direction from the ‘free’ tail
to the ‘free’ tip
8. Parallelogram of Vectors
Place the two vectors tail to tail, to scale and with the
correct directions
Then complete the parallelogram
The diagonal starting where the two tails meet and
finishing where the two arrows meet becomes the
resultant vector
10. More than 2
If there are more than 2 co-planar vectors to be added,
place them all head to tail to form polygon when the
resultant is drawn from the ‘free’ tail to the ‘free’ tip.
Notice that the order doesn’t matter!
11. EXAMPLE
A person walks 10 m due north and then 15 m due
west. Determine where the person is in relation to
where they started.
12. Subtraction of Vectors
To subtract a vector, you reverse the direction of that
vector to get the negative of it
Then you simply add that vector
14. Multiplying Scalars
Scalars are multiplied and divided in the normal
algebraic manner
Do not forget units!
5m / 2s = 2.5 ms-1
2kW x 3h = 6 kWh (kilowatt-hours)
15. Multiplying Vectors
A vector multiplied by a scalar gives a vector with the
same direction as the vector and magnitude equal to
the product of the scalar and a vector magnitude
A vector divided by a scalar gives a vector with same
direction as the vector and magnitude equal to the
vector magnitude divided by the scalar
You don’t need to be able to multiply a vector by
another vector
16. Resolving Vectors
The process of finding the Components of vectors is
called Resolving vectors
Just as 2 vectors can be added to give a resultant, a single
vector can be split into 2 components or parts
17. The Rule
A vector can be split into two perpendicular components
These could be the vertical and horizontal components
Vertical component
Horizontal component
19. These vertical and horizontal components could be the
vertical and horizontal components of velocity for
projectile motion
Or the forces perpendicular to and along an inclined
plane
20. Doing the Trigonometry
Sin = opp/hyp = y/V
Cos = adj/hyp = x/V
V
y
x
Therefore y = Vsin
In this case this is the
vertical component
Therefore x = Vcos
In this case this is the
horizontal component
V cos
V sin
21. Quick Way
If you resolve through the angle it is
cos
If you resolve ‘not’ through the angle it is
sin
22. Adding 2 or More Vectors by
Components
First resolve into components (making sure that all are
in the same 2 directions)
Then add the components in each of the 2 directions
Recombine them into a resultant vector
This will involve using Pythagoras´ theorem
23. Question
Three strings are attached to a small metal ring. 2 of
the strings make an angle of 70o and each is pulled
with a force of 7N.
What force must be applied to the 3rd string to keep
the ring stationary?
24. Answer
Draw a diagram
7N 7N
F
70o
7 sin 35o7 sin 35o
7 cos 35o + 7 cos 35o
25. Horizontally
7 sin 35o - 7 sin 35o = 0
Vertically
7 cos 35o + 7 cos 35o = F
F = 11.5N
And at what angle?
145o to one of the strings.
27. Kinematic Concepts
Kinematics is the part of mechanics that describes how
objects move.
The why objects move as they do is called dynamics.
28. Kinematic Concepts
Displacement
When a person moves over a time interval, they change
their position in space.
This change in position is called the:
displacement.
It is a vector quantity.
The further it moves from its starting position, the
greater it’s displacement.
29. Kinematic Concepts
If the object moves in the opposite direction;
as defined, its displacement will be negative.
31. Kinematic Concepts
The term distance is a scalar:
the symbol is d.
Distance is more useful when purchasing a car. The
distance travelled is important; while the direction it
travelled in is not.
32. Kinematic Concepts
A person who walks 100 m east and then 100 m west has
travelled a distance of, 200 m.
Their displacement however is 0 m.
They have ended up at their starting point.
Distance vs Displacement
33. Kinematic Concepts
Speed vs Velocity
Although speed and velocity are used interchangeably in
everyday life, both terms have specific meanings.
34. Kinematic Concepts
Speed is defined as
the distance travelled by an object in a given time
interval.
This will give us the average speed.
Mathematically, it can be represented by:
takentime
travelleddistancespeedaverage
35. Kinematic Concepts
As distance is one of the variables,
speed must be a scalar.
Velocity is used to signify both magnitude and direction
hence, it is a vector.
36. Kinematic Concepts
The average velocity is defined as:
the change in position of the object in a given time
interval.
Mathematically, it can be described as:
takentime
ntdisplacemelocityaverage ve
t
sv
37. Kinematic Concepts
The S.I. unit for both speed and velocity is m s-1.
Always include a direction when giving the value of the
velocity.
38. Kinematic Concepts
Acceleration.
An object whose velocity is changing is accelerating.
Acceleration is defined as the rate of change in velocity.
t
va D
39. Kinematic Concepts
Which cars below are accelerating and which
cars are traveling at constant velocity?
Which car accelerates fastest?
40. Kinematic Concepts
The red car is traveling at constant velocity.
The blue car is accelerating fastest. Its rate of change of
velocity is greatest.
41. Kinematic Concepts
Since an object always accelerates in a given direction,
acceleration is a vector quantity.
The S.I. units are m s-2
42. Kinematic Concepts
Changing Units
When using kinematic equations, the data given is not
always given in S.I. units.
The data needs to be converted to S.I. Units before they
can be substituted into an equation.
43. Kinematic Concepts
An example is speed.
Very often speed is given in km h-1.
The kilometres need to be converted to metres and the
hours into seconds.
There are 1000 m in 1 km and 3600 s in 1 hr (60 x 60).
1-1
sm7.27
60x60x1
1000x100
h1
km100
hkm100
44. Kinematic Concepts
There is doubt over whether the correct answer should
be:
30 m s-1,
28 m s-1 or,
27.7 m s-1
due to the number of significant figures given.
45. Kinematic Concepts
To avoid this problem, always give your data using
scientific notation.
In the above example,
1.00 x 102 m s-1 would eliminate the problem
hence the answer would be 27.7 m s-1.
46. Kinematic Concepts
Instantaneous vs Average
If you drive a car for 240 km in 3 hrs, your average speed
is 80 km h-1.
It is unlikely that for every part of the journey, you would
be travelling at 80 km h-1.
At each instant, your speed would change.
47. Kinematic Concepts
The speedometer in the car gives the instantaneous
speed.
The instantaneous speed is defined as:
the average speed over an indefinitely short time interval.
48. Kinematic Concepts
The same definition can be used for velocity.
The formula for velocity needs to be changed to
accommodate the difference between average, and
instantaneous velocity.
50. Kinematic Concepts
As Dt becomes very small,
approaching zero,
Ds approaches zero as well.
However, the ratio approaches a definite value.
51. Kinematic Concepts
This definite value is known as the instantaneous
velocity.
The symbol for instantaneous velocity is:
v no av or bar above the v.
tD
D
s
v
52. Kinematic Concepts
The same procedure can be used for acceleration and so
the equations become:
Average acceleration:
Instantaneous acceleration:
t
av
D
D
v
a
tD
D
v
a
_
tD
D
v
a
53. Graphical Representation of
Motion
Graphical Representation of Position
Multiflash photographs of an object in motion can be
taken and, data collected from it.
An example of a multiflash photo is the toy car on the
next slide.
54. Graphical Representation of
Motion
To take this photo;
the film needs to be exposed for a period of time in
darkness, with a strobe flashing at a known rate.
55. Graphical Representation of
Motion
From the data collected, graphs
can be drawn.
By plotting position on the
vertical axis and time on the
horizontal, the graph will look
like:
56. Graphical Representation of
Motion
A number of facts can be ascertained from this graph.
1. As the line is straight, the change in position per
unit time, is constant. This means the velocity is
constant.
2. The magnitude of the velocity can be obtained by
the slope
timeinchange
positioninchange
v
57. Graphical Representation of
Motion
If the graph is steep, it means:
there is a greater change in
position per unit time and the
object, is moving relatively
fast.
59. Graphical Representation of
Motion
If the graph is horizontal the object is stationary.
If the slope is negative the object is moving back
towards its starting position with, constant velocity.
61. Graphical Representation of
Motion
In graph a);
1. The person has moved at constant velocity over
the interval t1 to point p
2. They then remained stationary for the period t2
3. The person then returned to the original starting
position at a constant, but slower velocity over
the interval t3.
62. Graphical Representation of
Motion
In graph b);
1. The person has moved at constant velocity over
the interval t1 to point p
2. They then remained stationary for the period t2
3. The person then continued in the original
direction for the same distance p as in interval t1
at a constant, but slower velocity over the interval
t3.
63. Graphical Representation of
Motion
Graphical Representation of Velocity
Consider graph a) from above.
From the information from the graph, a graph of the
person’s velocity can be drawn.
65. Graphical Representation of
Motion
The graph is slightly idealised as the person could not
travel at the constant velocity, at every instant of the
journey.
66. Graphical Representation of
Motion
In a real situation,
velocity does not change in zero time and a more
likely description is, the one shown above.
70. Graphical Representation of
Motion
The distance from one image to the next is:
the change in position of the object in, a specific interval
of time.
Changes in position in equal intervals of time are:
direct measures of the average velocity, over those
intervals.
71. Graphical Representation of
Motion
The area under the Velocity vs. Time graph can be used
to obtain further information.
This information relates to the change in position of the
object.
72. Graphical Representation of
Motion
Rearranging this equation:
change in position is given by the product;
average velocity x time taken
The graph on the next slide deals with constant velocity.
timeinchange
positioninchange
v
_
74. Graphical Representation of
Motion
As the velocity is constant the average velocity must be v.
Hence, the change in position will be vt.
This is also equal to the area under graph in time interval
t.
This relationship holds for all v/t graphs as in the
example on the next slide:
76. Graphical Representation of
Motion
The motion in the graph above describes:
motion under non-zero constant acceleration eg
motion under the force of gravity.
77. Graphical Representation of
Motion
Since velocity changes in a regular way the average
velocity is,
the average of the initial velocity and,
the final velocity.
The horizontal line through P represents this.
80. Graphical Representation of
Motion
Another way to calculate the change in position is:
to divide the area under the graph into, a rectangle and
triangle and, add the two solutions.
81. Graphical Representation of
Motion
Any area below the axis corresponds to:
motion in the opposite direction where, the change in
position is opposite in sign.
83. Graphical Representation of
Motion
In the example, the
slope, and therefore the
rate at which velocity is
changing, is constant.
Constant Acceleration
Graph
Time (sec)
Velocity
(ms -1)
85. Graphical Representation of
Motion
The slope of the graph is given by: rise/run.
In the example, the rise = Dv and the run as Dt.
Time (sec)
Velocity
(ms -1)
87. Graphical Representation of
Motion
This corresponds to our formula for acceleration.
The slope of a v/t graph gives:
acceleration.
Acceleration is measured in m s-2 and is a vector quantity
so, direction must always be included.
88. Graphical Representation of
Motion
A car is stationary at the lights when the lights change to
green.
Another car is moving when the lights turn green.
What is the displacement of each car after 3 seconds?
93. Graphical Representation of
Motion
Slope = rise/run
Acceleration = 12/3
Acceleration = 4 m s-2
Watch the animation again and determine the
displacement of both cars after 9 seconds.
95. Graphical Representation of
Motion
Red Car
Displacement = area of triangle + area of rectangle.
Displacement = (½ x 3 x 12) + (9 x 12)
Displacement =18 + 72
Displacement = 90 m
96. Graphical Representation of
Motion
Blue Car
Displacement = area of rectangle
Displacement = 9 x 10
Displacement = 90 m
Watch the animation again.
When do the two cars pass each other?
Does it agree with your calculations?
104. Graphical Representation of
Motion
Very complicated motion can be studied using graphs.
Watch the two stage rocket as it is launched, run out of
fuel and returns to Earth.
106. Graphical Representation of
Motion
The Moving Man
Motion Graphs
If you would like more practice at drawing motion
graphs diagrams, try this web site:
http://www.glenbrook.k12.il.us/gbssci/phys/shwave/gra
ph.html
107. Equations for Uniformly
Accelerated Motion
The previous equations used can be applied to all types
of motion.
However, when acceleration is constant our
mathematical description can be taken further.
109. Equations for Uniformly
Accelerated Motion
In the graph on the previous slide, as the slope is
constant the acceleration is also constant.
If we let u and v be the velocity at the start and end of
the time interval t average velocity can be described as;
2
vu
vav
111. Equations for Uniformly
Accelerated Motion
The average acceleration can also be given
as:
Rearranging the equation gives:
t
uv
a
atuv
112. Equations for Uniformly
Accelerated Motion
These two equations describe motion:
at constant acceleration,
in terms of five variables;
u, v, t, s, a.
113. Equations for Uniformly
Accelerated Motion
Equation u does not use a and v does not use s.
Using algebra, we can derive 3 more equations, each one
not using one variable listed above.
120. Equations for Uniformly
Accelerated Motion
u v s
t
2
s ut at
1
2
2
s vt at
1
2
2
VARIABLES EQUATION
u v t s a
u v t s a v = u + at
u v t s a v2 - u2 = 2as
u v t s a
u v t s a
TOK
Physicists reduce concepts to
mathematical equations to reduce
ambiguity. Does mathematics and
equations reduce ambiguity due to
language?
TOK
121. Vertical Motion of Objects
The Earth, near its surface, has a uniform gravitational
field.
Objects that move perpendicular to the Earth’s surface,
move parallel to the gravitational field.
122. Vertical Motion of Objects
The acceleration experienced by the object will be
constant.
a = 9.81 m s-2 down.
If the object is moving towards the Earth, the object’s
speed will increase at the rate of, 9.81 m s-1 every second.
This is assuming that air resistance is negligible.
124. Vertical Motion of Objects
True or False?
1. The elephant encounters a smaller force
of air resistance than the feather and
therefore falls faster.
2. The elephant has a greater acceleration
of gravity than the feather and therefore
falls faster.
TOK
125. Vertical Motion of Objects
3. Both elephant and feather have the same
force of gravity, yet the acceleration of
gravity is greatest for the elephant.
4. Both elephant and feather have the same
force of gravity, yet the feather experiences
a greater air resistance.
TOK
126. Vertical Motion of Objects
5. Each object experiences the same amount
of air resistance, yet the elephant
experiences the greatest force of gravity.
6. Each object experiences the same amount
of air resistance, yet the feather
experiences the greatest force of gravity.
TOK
127. Vertical Motion of Objects
7. The feather weighs more than the elephant,
and therefore will not accelerate as rapidly
as the elephant.
8. Both elephant and feather weigh the same
amount, yet the greater mass of the feather
leads to a smaller acceleration.
TOK
128. Vertical Motion of Objects
9. The elephant experiences less air resistance
and than the feather and thus reaches a
larger terminal velocity.
10.The feather experiences more air resistance
than the elephant and thus reaches a
smaller terminal velocity.
TOK
129. Vertical Motion of Objects
11.The elephant and the feather encounter the
same amount of air resistance, yet the
elephant has a greater terminal velocity.
If you answered True to any of these questions,
you need to review your understanding.
Consider, how you knew?
TOK
131. Vertical Motion of Objects
Objects which move vertically upwards will slow down at
the rate of 9.81 m s-1, every second until it is stationary.
It will then start to accelerate towards the earth at the
rate of 9.81 m s-1 every second.
Vertical Motion
132. Air Resistance
Objects falling in the Earth’s uniform gravitational field
have two opposing forces acting on it.
Gravity acts towards the Earth, pulling the object
downward.
Any resistance force opposes motion.
133. Air Resistance
This means, in this case bair resistance acts upwards.
The faster the object falls the greater the air resistance.
As the object accelerates under the force of gravity the
greater the air resistance.
134. Air Resistance
This slows the rate at which the object accelerates
towards the earth.
Eventually, the two forces cancel each other out.
This means there is no acceleration and the velocity
becomes constant.
This is known as terminal velocity.
135. Air Resistance
The terminal velocity is different for different objects.
Sky diver’s have a terminal velocity of :
about 150 to 200 km h-1 depending on;
the mass of the sky diver and,
their orientation.
136. Air Resistance
Sky divers tend to try to:
increase the air resistance thereby,
reducing the terminal velocity.
This gives them a longer free fall.
The parachute:
greatly increases air resistance and,
cuts the terminal velocity to,
between 15 and 20 km h-1.
138. Components of Motion
When a body is in free motion, (moving through the
air without any forces apart from gravity and air
resistance), it is called a projectile
Normally air resistance is ignored so the only force
acting on the object is the force due to gravity
This is a uniform force acting downwards
139. Therefore if the motion of the projectile is resolved
into the vertical and horizontal components
The horizontal component will be unaffected as
there are no forces acting on it
The vertical component will be accelerated
downwards by the force due to gravity
140. These two components can be considered as
independent factors in the motion of a projectile in a
uniform field
In the absence of air resistance the path taken by any
projectile is parabolic
141. Solving Problems
In solving problems it is necessary to consider the 2
components independently
142. Therefore the horizontal motion it is necessary to use
the equation
speed = distance
time
Where speed is the horizontal component of the velocity
143. Therefore the vertical motion it is necessary to use the
kinematic equations for uniform acceleration
i.e. Using the s.u.v.a.t equations
Where u and v are the initial and final vertical
components of the velocity
144. Example
A ball is kicked at an angle of 40.0o with a
velocity of 10.0 ms-1. Taking g = 10 ms–2. How far
does it travel horizontally?
40o
10ms-1
145. To be able to calculate the horizontal distance we
need to know the horizontal speed, and the time.
The horizontal distance is easy to calculate by
resolving the velocity
40.0o
10.0ms-110.0 sin 40.0o
10.0 cos 40.0o
146. However, to calculate the time we will need to use the
vertical component and the s.u.v.a.t. Equations
147. s = ?
u = 10.0 sin 40.0o ms-1
v = ?
a = -10.0 ms-2 (Up is positive, therefore acceleration
here is negative)
t = ?
We only have 2 of the values when we need three to find
any other
148. However, if we ignore air resistance, then the final
vertical component of the velocity will be equal and
opposite of the initial component
i.e. v = -10.0 sin 40.0o ms-1
Looking at the equations for uniform acceleration, we
need an equation that links u, v, a and t.
149. v = u + at
Rearranging to make t the subject
t = v – u a
Substitute in
t = -10.0 sin 40.0o – 10.0 sin 40.0o
-10
t = 1.286 seconds
150. Now returning to the horizontal components
Using speed = distance
time
Rearranging distance = speed x time
Distance = 10.0 cos 40.0o x 1.286
Distance = 9.851 = 9.9 metres
151. Using the Conservation of Energy
In some situations the use of the conservation of
energy can be a much simpler method than using
the kinematic equations
Solving projectile motion problems makes use of the
fact that Ek + Ep = constant at every point in the
objects flight (assuming no loss of energy due to
friction)
152. Example
A ball is projected at 25.0 ms-1 at an angle of 40.00 to
the horizontal. The ball is released 2.00m above the
ground. Taking g = 10.0 ms-2. Find the maximum
height it reaches.
154. Total energy at A is given by
Ek + Ep = ½ m (25.0)2 + mg x 2.0
=312.5m + 20m
= 332.5m
155. Next, to find the total energy at B we need to know the
velocity at B, which is given by the horizontal
component of the velocity at A
Total energy at B is given by
Ek+ Ep = ½ m (25.0 cos 40o)2 + mg x H
=183.38m+ 10mH
Then using the conservation of energy
